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| | <span class="exam">(b) Use the Intermediate Value Theorem to show that <math style="vertical-align: -5px">f(x)</math> has a zero in the interval <math style="vertical-align: -5px">[0,1].</math> | | <span class="exam">(b) Use the Intermediate Value Theorem to show that <math style="vertical-align: -5px">f(x)</math> has a zero in the interval <math style="vertical-align: -5px">[0,1].</math> |
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| | + | [[009A Sample Midterm 2, Problem 2 Detailed Solution|'''<u>Detailed Solution with Background Information</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[File:9ASM2P2.jpg|600px|thumb|center]] |
| − | !Foundations:
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| − | |What is a zero of the function <math style="vertical-align: -5px">f(x)?</math> | |
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| − | | A zero is a value <math style="vertical-align: -1px">c</math> such that <math style="vertical-align: -5px">f(c)=0.</math>
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| − | |} | |
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| − | '''Solution:'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !(a)
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| − | |-
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| − | |'''Intermediate Value Theorem'''
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| − | | If <math style="vertical-align: -5px">f(x)</math> is continuous on a closed interval <math style="vertical-align: -5px">[a,b]</math>
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| − | |-
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| − | | and <math style="vertical-align: 0px">c</math> is any number between <math style="vertical-align: -5px">f(a)</math> and <math style="vertical-align: -5px">f(b),</math>
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| − | |-
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| − | |
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| − | then there is at least one number <math style="vertical-align: 0px">x</math> in the closed interval such that <math style="vertical-align: -5px">f(x)=c.</math>
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| − | |}
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| − |
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| − | '''(b)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |First, <math style="vertical-align: -5px">f(x)</math> is continuous on the interval <math style="vertical-align: -5px">[0,1]</math> since <math style="vertical-align: -5px">f(x)</math> is continuous everywhere.
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| − | |-
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| − | |Also,
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| − | <math>f(0)=2</math>
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| − | |-
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| − | |and
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| − | <math>f(1)=3-8+2=-3.</math>.
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Since <math style="vertical-align: -5px">y=0</math> is between <math style="vertical-align: -5px">f(0)=2</math> and <math style="vertical-align: -5px">f(1)=-3,</math>
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| − | |-
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| − | |the Intermediate Value Theorem tells us that there is at least one number <math style="vertical-align: -1px">x</math>
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| − | |-
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| − | |such that <math style="vertical-align: -5px">f(x)=0.</math>
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| − | |-
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| − | |This means that <math style="vertical-align: -5px">f(x)</math> has a zero in the interval <math style="vertical-align: -5px">[0,1].</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | '''(a)''' See solution above.
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| − | |-
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| − | | '''(b)''' See solution above.
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| − | |}
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| | [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] |
The function 
is a polynomial and therefore continuous everywhere.
(a) State the Intermediate Value Theorem.
(b) Use the Intermediate Value Theorem to show that 
has a zero in the interval ![{\displaystyle [0,1].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8786b5ef9daedb24adb59e7825c4096d99a99648)
Detailed Solution with Background Information
Return to Sample Exam
